// 代码题三、基于下面提供的代码，完成后续的四个练习

const fp = require('lodash/fp')
const { Maybe, Container } = require('./support')
const { parseInt } = require('lodash')


// 练习1: 使用fp.add(x, y)和fp.map(f, x)创建一个能让functor里的值增加的函数ex1
let maybe = Maybe.of([5, 6, 1])
let ex1= (functor, val) => {
    functor._value = fp.map(fp.add(val), functor._value)
}
ex1(maybe, 2)
console.log(maybe._value)


// 练习2: 实现一个函数ex2,能够使用fp.first获取列表的第一个元素
let xs = Container.of (['do', 'ray','me', 'fa', 'so', 'la', 'ti', 'do'])
let ex2 = container => {
    return fp.first(container._value)
}
console.log(ex2(xs))


// 练习3: 实现一个函数ex3,使用safeProp和fp.first找到user的名字的首字母
let safeProp = fp.curry(function (x,o) {
    return Maybe.of(o[x])
})
let user = { id: 2, name: 'Albert'}
let ex3 = user => {
    return fp.first(safeProp('name', user)._value)
}
console.log(ex3(user))


// 练习4: 使用Maybe重写ex4,不要有if语句
let ex4 = num => {
    let maybe = Maybe.of(num)
    return maybe.map(parseInt)._value
}
console.log(ex4('33.33'))